Array remove duplicate elements

Question

I have an unsorted array, what is the best method to remove all the duplicates of an element if present?

e.g:

a[1,5,2,6,8,9,1,1,10,3,2,4,1,3,11,3]


        

Answer 1

Treat numbers as keys.

for each elem in array:
if hash(elem) == 1 //duplicate
  ignore it
  next
else
  hash(elem) = 1
  add this to resulting array 
end

If you know about the data like the range of numbers and if it is finite, then you can initialize that big array with ZERO's.

array flag[N] //N is the max number in the array
for each elem in input array:
  if flag[elem - 1] == 0
    flag[elem - 1] = 1
    add it to resulatant array
  else
    discard it //duplicate
  end

Answer 2

You can use the "in" and "not in" syntax in python which makes it pretty straight forward.

The complexity is higher than the hashing approach though since a "not in" is equivalent to a linear traversal to find out whether that entry exists or not.

li = map(int, raw_input().split(","))
a = []
for i in li:
    if i not in a:
        a.append(i)
print a

Answer 3

If you don't need to keep the original object you can loop it and create a new array of unique values. In C# use a List to get access to the required functionality. It's not the most attractive or intelligent solution, but it works.

int[] numbers = new int[] {1,2,3,4,5,1,2,2,2,3,4,5,5,5,5,4,3,2,3,4,5};
List<int> unique = new List<int>();

foreach (int i in numbers)
     if (!unique.Contains(i))
          unique.Add(i);

unique.Sort();
numbers = unique.ToArray();

Answer 4

    indexOutput = 1;
    outputArray[0] = arrayInt[0];
    int j;
    for (int i = 1; i < arrayInt.length; i++) {            
        j = 0;
        while ((outputArray[j] != arrayInt[i]) && j < indexOutput) {
            j++;
        }
        if(j == indexOutput){
           outputArray[indexOutput] = arrayInt[i];
           indexOutput++;
        }         
    }

Answer 5

This is a code segment i created in C++, Try out it

#include <iostream>

using namespace std;

int main()
{
   cout << " Delete the duplicate" << endl; 

   int numberOfLoop = 10;
   int loopCount =0;
   int indexOfLargeNumber = 0;
   int largeValue = 0;
   int indexOutput = 1;

   //Array to hold the numbers
   int arrayInt[10] = {};
   int outputArray [10] = {};

   // Loop for reading the numbers from the user input
   while(loopCount < numberOfLoop){       
       cout << "Please enter one Integer number" << endl;
       cin  >> arrayInt[loopCount];
       loopCount = loopCount + 1;
   }



    outputArray[0] = arrayInt[0];
    int j;
    for (int i = 1; i < numberOfLoop; i++) {            
        j = 0;
        while ((outputArray[j] != arrayInt[i]) && j < indexOutput) {
            j++;
        }
        if(j == indexOutput){
           outputArray[indexOutput] = arrayInt[i];
           indexOutput++;
        }         
    }

   cout << "Printing the Non duplicate array"<< endl;

   //Reset the loop count
   loopCount =0;

   while(loopCount < numberOfLoop){ 
       if(outputArray[loopCount] != 0){
        cout <<  outputArray[loopCount] << endl;
    }     

       loopCount = loopCount + 1;
   }   
   return 0;
}

Answer 6

Time O(n) space O(n) 

#include <iostream>
    #include<limits.h>
    using namespace std;
    void fun(int arr[],int size){

        int count=0;
        int has[100]={0};
        for(int i=0;i<size;i++){
            if(!has[arr[i]]){
               arr[count++]=arr[i];
               has[arr[i]]=1;
            }
        }
     for(int i=0;i<count;i++)
       cout<<arr[i]<<" ";
    }

    int main()
    {
        //cout << "Hello World!" << endl;
        int arr[]={4, 8, 4, 1, 1, 2, 9};
        int size=sizeof(arr)/sizeof(arr[0]);
        fun(arr,size);

        return 0;
    }